CodeForces 337C 找规律 ,等比数列求和

H - 1007

Time Limit:1000MS    Memory Limit:262144KB    64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 337C

Description

Manao is taking part in a quiz. The quiz consists of n consecutive questions. A correct answer gives one point to the player. The game also has a counter of consecutive correct answers. When the player answers a question correctly, the number on this counter increases by 1. If the player answers a question incorrectly, the counter is reset, that is, the number on it reduces to 0. If after an answer the counter reaches the number k, then it is reset, and the player's score is doubled. Note that in this case, first 1 point is added to the player's score, and then the total score is doubled. At the beginning of the game, both the player's score and the counter of consecutive correct answers are set to zero.

Manao remembers that he has answered exactly m questions correctly. But he does not remember the order in which the questions came. He's trying to figure out what his minimum score may be. Help him and compute the remainder of the corresponding number after division by 1000000009(10⁹ + 9).

Input

The single line contains three space-separated integers n, m and k (2 ≤ k ≤ n ≤ 10⁹; 0 ≤ m ≤ n).

Output

Print a single integer — the remainder from division of Manao's minimum possible score in the quiz by 1000000009(10⁹ + 9).

Sample Input

Input

5 3 2

Output

3

Input

5 4 2

Output

6

Hint

Sample 1. Manao answered 3 questions out of 5, and his score would double for each two consecutive correct answers. If Manao had answered the first, third and fifth questions, he would have scored as much as 3 points.

Sample 2. Now Manao answered 4 questions. The minimum possible score is obtained when the only wrong answer is to the question 4.

Also note that you are asked to minimize the score and not the remainder of the score modulo 1000000009. For example, if Manao could obtain either 2000000000 or 2000000020 points, the answer is 2000000000 mod 1000000009, even though 2000000020 mod 1000000009 is a smaller number.

题意是n 个题,答对m个题,然后是如果有k个连续的题答对了就让总和加倍

满足sum=(sum+k)*2;

//为了使得分最小,先让n段数里隔一个存k-1个数
//多余的数从前往后存,填上间隔,

于是推出了sum=(sum+k)*2;的公式TLE

((2*k+k)*2+k)*2;

提取别先合并,

一整合就是等比数列求和公式,用快速幂

2 6 14 30 62 126..等比求和

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <cmath>
using namespace std;
long long Pow(long long int a,int b)
{
    long long sum=1;
    while(b)
    {
        if(b&1)
            sum=a*sum%1000000009;
        b/=2;
        a=a*a%1000000009;
    }
    return sum;
}
int main()  
{
    long long n,m,k;
    long long x,y,z,num,sum;
    while(~scanf("%lld%lld%lld",&n,&m,&k))
    {
        sum=0; //
        x=n-n/k; //不加倍能够 存的最大得分
        num=m-x;   //需要加倍的次数
        if(num<=0)
        {
            printf("%lld\n",m%1000000009);
            continue;
        }
        sum=(Pow(2,num)*k-k)*2%1000000009;
        num=m-k*num%1000000009;
        printf("%lld\n",(num+sum)%1000000009);
    }
    return 0;
}